The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double-slit experiment,where the phase differences between two waves of the same amplitude are $\pi / 3$ and $\pi / 2$ respectively,is:

In Young's double slit experiment,the light emitted from the source has $\lambda = 6.5 \times 10^{-7} \, m$ and the distance between the two slits is $1 \, mm$. The distance between the screen and the slits is $1 \, m$. The distance between the third dark fringe and the fifth bright fringe will be ......... $mm$.

In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then:

In the figure, Young's double-slit experiment is shown. $Q$ is the position of the first bright fringe on the right side of $O$. $P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{-10} \text{ m}$, then $S_1B$ will be equal to:

In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:

In Young's double-slit experiment,when light of wavelength $\lambda_1 = 5898 \ \mathring{A}$ is used,$92$ fringes are observed in a certain region. If light of wavelength $\lambda_2 = 5461 \ \mathring{A}$ is used instead,how many fringes will be observed in the same region?